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x-3-5-2-3-y-3-5-2-3-x-




Question Number 193410 by cortano12 last updated on 13/Jun/23
   { ((x=(√(3−(√(5+2(√3))))))),((y=(√(3+(√(5+2(√3))))))) :}       x
$$\:\:\begin{cases}{\mathrm{x}=\sqrt{\mathrm{3}−\sqrt{\mathrm{5}+\mathrm{2}\sqrt{\mathrm{3}}}}}\\{\mathrm{y}=\sqrt{\mathrm{3}+\sqrt{\mathrm{5}+\mathrm{2}\sqrt{\mathrm{3}}}}}\end{cases}\: \\ $$$$\:\:\:\:\underbrace{\boldsymbol{{x}}} \\ $$
Answered by aba last updated on 13/Jun/23
xy=(√(9−(5+2(√3))))=(√(4−2(√3)))=(√3)−1 ∧ x^2 +y^2 =6  x+y=(√(x^2 +y^2 +2xy))=(√(6+2((√3)−1)))=(√(4+2(√3)))=(√3)+1
$$\mathrm{xy}=\sqrt{\mathrm{9}−\left(\mathrm{5}+\mathrm{2}\sqrt{\mathrm{3}}\right)}=\sqrt{\mathrm{4}−\mathrm{2}\sqrt{\mathrm{3}}}=\sqrt{\mathrm{3}}−\mathrm{1}\:\wedge\:\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} =\mathrm{6} \\ $$$$\mathrm{x}+\mathrm{y}=\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} +\mathrm{2xy}}=\sqrt{\mathrm{6}+\mathrm{2}\left(\sqrt{\mathrm{3}}−\mathrm{1}\right)}=\sqrt{\mathrm{4}+\mathrm{2}\sqrt{\mathrm{3}}}=\sqrt{\mathrm{3}}+\mathrm{1} \\ $$
Answered by Tinku Tara last updated on 13/Jun/23
(x+y)^2 =x^2 +y^2 +2xy  =6+2(√(9−5−2(√3)))  =6+2(√(4−2(√3)))     (4−2(√3)=((√3)−(√1))^2 )  =6+2((√3)−1)  =4+2(√3)  =((√3)+(√1))^2   x+y=1+(√3)
$$\left({x}+{y}\right)^{\mathrm{2}} ={x}^{\mathrm{2}} +{y}^{\mathrm{2}} +\mathrm{2}{xy} \\ $$$$=\mathrm{6}+\mathrm{2}\sqrt{\mathrm{9}−\mathrm{5}−\mathrm{2}\sqrt{\mathrm{3}}} \\ $$$$=\mathrm{6}+\mathrm{2}\sqrt{\mathrm{4}−\mathrm{2}\sqrt{\mathrm{3}}}\:\:\:\:\:\left(\mathrm{4}−\mathrm{2}\sqrt{\mathrm{3}}=\left(\sqrt{\mathrm{3}}−\sqrt{\mathrm{1}}\right)^{\mathrm{2}} \right) \\ $$$$=\mathrm{6}+\mathrm{2}\left(\sqrt{\mathrm{3}}−\mathrm{1}\right) \\ $$$$=\mathrm{4}+\mathrm{2}\sqrt{\mathrm{3}} \\ $$$$=\left(\sqrt{\mathrm{3}}+\sqrt{\mathrm{1}}\right)^{\mathrm{2}} \\ $$$${x}+{y}=\mathrm{1}+\sqrt{\mathrm{3}} \\ $$

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